02b_Reinforced and Prestressed Concrete_Engineering and Design

5 Calculate the moment of inertia of the transformed cracked section at the center Referring to Fig I9e and assuming tentatively that the neutral axis falls within the flange, we see nAs = 10(3.16) = 31.6 in2 (203.88 cm2) The static moment with respect to the neutral axis is Q = l/2(68y2) - 31.6(20.5 -y) = O; y = 3.92 in (99.568 mm) The neutral axis therefore falls within the flange, as assumed The moment of inertia with respect to the neutral axis is I2 = 0/3)68(3.92)3 + 31.6(20.5 - 3.92)2 = 10,052 in4 (41.840 dm4) Calculate the deflection at midspan Use the equation A= L'2 i 5M1 M3 \ -Ji (IT-T) (37) where = average moment of inertia, in4 (dm4) Thus, / = ^(9737 + 10,052) = 9895 in4 (41.186 dm4); E = 14515 x 33/c')°5 = 57,600(250O)05 = 2,880,000 lb/in2 (19,857.6 MPa) Then A = [222 x 17287(2880 x 9895)](5 x 200/48 - 100/8) = 0.244 in (6.198 mm) Where the deflection under sustained loading is to be evaluated, it is necessary to apply the factors recorded in the ACI Code Design of Compression Members by Ultimate-Strength Method The notational system is Pu = ultimate axial compressive load on member, Ib (N); Pb = ultimate axial compressive load at balanced design, Ib (N); P0 = allowable ultimate axial compressive load in absence of bending moment, Ib (N); Mu = ultimate bending moment in member, lb-in (N-m); Mb = ultimate bending moment at balanced design; d' = distance from exterior surface to centroidal axis of adjacent row of steel bars, in (mm); t = overall depth of rectangular section or diameter of circular section, in (mm) A compression member is said to be spirally reinforced if the longitudinal reinforcement is held in position by spiral hooping and tied if this reinforcement is held by means of intermittent lateral ties The presence of a bending moment in a compression member reduces the ultimate axial load that the member may carry In compliance with the ACI Code, it is necessary to design for a minimum bending moment equal to that caused by an eccentricity of 0.05/ for spirally reinforced members and 0.10/ for tied members Thus, every compression member that is designed by the ultimate-strength method must be treated as a beam column This type of member is considered to be in balanced design if failure would be characterized by the simultaneous crushing of the concrete, which is assumed to occur when ec = 0.003, and incipient yielding of the tension steel, which occurs when^ =fy The ACI Code set = 0.75 for spirally reinforced members and 1; therefore,/^ =£; ab = 0.85(10.62) = 9.03 in (229.362 mm); F0 = 0.85(3000)(12^) = 276,300 Ib (1,228,982.4 N); FA = 40,000(2.00) = 80,000 Ib ((355,84O N); F3 = - 80,000 Ib (-355,840 N); Pb = 0.70(276,300) = 193,400 Ib (860,243.2 N) Also, r Fc(t - d) (FA - FB)(t -2d'}-\ Mb = 0.70 I -^-^ + — y -I (38) Thus, Mb = 0.70(276,300(18 - 9.03)72 + 160,000(6.5)] = 1,596,000 in-lb (180,316.1 N-m) When c > cb9 the member fails by crushing of the concrete; when c < cb, it fails by yielding of the reinforcement at line B Compute the value of c associated with incipient yielding of the compression steel Compute the corresponding values of P11 and M14 Since eA and es are numerically equal, the neutral axis lies at N Thus, c = in (228.6 mm); a = 0.85(9) = 7.65 in (194.31 mm); F0 = 30,600(7.65) = 234,100 Ib (1,041,276.8 N); FA = 80,000 Ib (355,840 N); FB = -80,000 Ib (- 355,840 N); Pn = 0.70 (234,100) = 163,900 Ib (729,027.2 N); Mu = 0.70(234,100 x 5.18 + 160,000 x 6.5) = 1,577,000 in-lb (178,169.5 N-m) Compute the minimum value of c at which the entire concrete area is stressed to 085 f^ Compute the corresponding values of P11 and Mu Thus, a = t = 18 in (457.2 mm); c = 18/0.85 = 21.8 in (537.972 mm);/* = €